Motion Analysis: Velocity And Acceleration Calculation

Hey guys! Let's dive into a classic problem from kinematics: analyzing the motion of a particle along a straight line. We've got a particle whose position is described by the function $s(t) = 2t^3 - 7t^2 - 7t + 15$, where $s$ is measured in meters and $t$ in seconds, with time ($t$) being non-negative. Our mission is to find the velocity and acceleration functions of this particle. Buckle up, because it's going to be a fun ride through calculus!

Understanding the Problem: Position, Velocity, and Acceleration

Before we jump into calculations, let's make sure we're all on the same page with the core concepts. You see, the position function, $s(t)$, tells us where the particle is located at any given time, $t$. But to truly understand the particle's motion, we need to know not just where it is, but also how fast it's moving and how its speed is changing. That's where velocity and acceleration come into play.

Velocity, often denoted as $v(t)$, is the rate of change of position with respect to time. In simpler terms, it tells us how quickly the particle is moving and in what direction. If the velocity is positive, the particle is moving in the positive direction; if it's negative, the particle is moving in the negative direction; and if it's zero, the particle is momentarily at rest. Velocity is a vector quantity, meaning it has both magnitude (speed) and direction.

Now, what about acceleration? Acceleration, usually written as $a(t)$, is the rate of change of velocity with respect to time. Think of it as how quickly the particle's velocity is changing. If the acceleration is positive, the velocity is increasing (the particle is speeding up in the positive direction or slowing down in the negative direction); if it's negative, the velocity is decreasing (the particle is slowing down in the positive direction or speeding up in the negative direction); and if it's zero, the velocity is constant. Like velocity, acceleration is also a vector quantity.

In the language of calculus, velocity is the first derivative of the position function, and acceleration is the first derivative of the velocity function (or, equivalently, the second derivative of the position function). So, to find $v(t)$ and $a(t)$, we'll need to flex our differentiation muscles!

Finding the Velocity Function

Okay, let's get our hands dirty with some math! We're given the position function:

$s(t) = 2t^3 - 7t^2 - 7t + 15$

To find the velocity function, $v(t)$, we need to take the first derivative of $s(t)$ with respect to time, $t$. Remember the power rule for differentiation: if we have a term of the form $at^n$, its derivative is $nat^{n-1}$. Applying this rule to each term in $s(t)$, we get:

$v(t) = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 7t^2 - 7t + 15)$

$v(t) = 2(3t^2) - 7(2t) - 7(1) + 0$

$v(t) = 6t^2 - 14t - 7$

So, there you have it! The velocity function for our particle is $v(t) = 6t^2 - 14t - 7$ meters per second. This equation tells us the particle's instantaneous velocity at any time, $t$.

Determining the Acceleration Function

Next up, let's find the acceleration function, $a(t)$. As we discussed, acceleration is the rate of change of velocity, so we need to take the derivative of the velocity function, $v(t)$, with respect to time:

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 14t - 7)$

Again, we'll use the power rule for differentiation:

$a(t) = 6(2t) - 14(1) - 0$

$a(t) = 12t - 14$

And voilà! The acceleration function for the particle is $a(t) = 12t - 14$ meters per second squared. This function tells us how the particle's velocity is changing at any given time, $t$.

Putting It All Together: Interpreting the Results

We've successfully found both the velocity and acceleration functions. Let's recap:

• Position function: $s(t) = 2t^3 - 7t^2 - 7t + 15$ (meters)
• Velocity function: $v(t) = 6t^2 - 14t - 7$ (meters per second)
• Acceleration function: $a(t) = 12t - 14$ (meters per second squared)

Now, what can we do with these functions? Well, a lot! We can analyze the particle's motion in detail. For instance, we could:

• Find the velocity and acceleration at a specific time by plugging in a value for $t$.
• Determine when the particle is at rest by setting $v(t) = 0$ and solving for $t$.
• Find the intervals where the particle is speeding up or slowing down by analyzing the signs of $v(t)$ and $a(t)$.
• Determine the particle's displacement and total distance traveled over a given time interval.

This is just the tip of the iceberg. The beauty of having these functions is that they provide a complete mathematical description of the particle's motion, allowing us to answer a wide range of questions about its behavior.

An Example: Finding Velocity and Acceleration at t = 2 seconds

To illustrate, let's find the velocity and acceleration of the particle at $t = 2$ seconds. We simply plug $t = 2$ into our velocity and acceleration functions:

$v(2) = 6(2)^2 - 14(2) - 7 = 6(4) - 28 - 7 = 24 - 28 - 7 = -11$ meters per second

$a(2) = 12(2) - 14 = 24 - 14 = 10$ meters per second squared

So, at $t = 2$ seconds, the particle has a velocity of -11 meters per second (meaning it's moving in the negative direction) and an acceleration of 10 meters per second squared (meaning its velocity is increasing in the positive direction, or becoming less negative).

Conclusion: The Power of Calculus in Kinematics

There you have it, folks! By applying the principles of calculus, specifically differentiation, we've successfully determined the velocity and acceleration functions for a particle moving along a straight line. This example showcases the immense power of calculus in analyzing motion and understanding the relationships between position, velocity, and acceleration. Remember, velocity is the first derivative of position, and acceleration is the first derivative of velocity (or the second derivative of position). With these tools in your arsenal, you're well-equipped to tackle a wide range of kinematics problems and unravel the mysteries of motion!

Keep practicing, keep exploring, and keep your curiosity alive. Until next time, happy problem-solving!